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EMA 4314
HW7 solutions
20.22 (a)
D
=
1
3
λ
c
'
=
kT
σ
P
c
'
=
8
RT
π
M
eq. 20.22, 20.13, 20.7
a.
D
=
(
1
3
)(
(1.38
×
10
−
23
J
K
)(298
K
)
(3.6
×
10
−
19
m
2
)(1
Pa
)
(8*8.3145
J
K
*
mol
)(298
K
)
* 39.948
×
10
−
3
kg
mol
=
1.51
m
2
s
This value differs from the book by a factor of 2/3, which is explained on page 775.
b. D = 1.51x10
5
m
2
s
1
c. D = 1.51x10
7
m
2
s
1
J
=
D
•
N
dP
dx
N
=
P
RT
dP
dx
=
0.1
atm
cm
=
1.013
×
10
6
Pa
m
a. J = 617 mol/m
2
*s
b. J = 617 x 10
3
mol/m
2
*s
c. J = 617 x 10
7
mol/m
2
*s
20.30 (a)
<
x
2
>
1/ 2
=
Dt
t
=
<
x
2
>
2
D
t
=
(5.0
×
10
−
3
m
)
2
2(3.17
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This note was uploaded on 11/16/2011 for the course EMA 4314 taught by Professor Phillpot during the Fall '10 term at University of Florida.
 Fall '10
 Phillpot

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